Oxidation number of sulphur in `Na_(2)S_(2)O_(3)` is

Options

A. `+2`
B. `+4`
C. `-2`
D. 0

Answer

To determine the oxidation number of sulfur in the compound \( \text{Na}_2\text{S}_2\text{O}_3 \), we can follow these steps:

### Step 1: Identify the oxidation states of known elements
- Sodium (Na) is an alkali metal, and its oxidation state is always +1.
- Oxygen (O) typically has an oxidation state of -2.

### Step 2: Write the formula for the compound
The compound is \( \text{Na}_2\text{S}_2\text{O}_3 \).

### Step 3: Set up the equation based on the oxidation states
Let the oxidation state of sulfur (S) be \( x \). In the compound, we have:
- 2 sodium atoms contributing \( 2 \times (+1) = +2 \)
- 3 oxygen atoms contributing \( 3 \times (-2) = -6 \)
- 2 sulfur atoms contributing \( 2x \)

The overall charge of the compound is neutral, which means the sum of the oxidation states must equal 0.

### Step 4: Write the equation
The equation can be set up as follows:
\[ 2 + 2x - 6 = 0 \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ 2x - 4 = 0 \]
\[ 2x = 4 \]
\[ x = 2 \]

### Step 6: Determine the oxidation states of the two sulfur atoms
Since we have two sulfur atoms, we need to consider their individual oxidation states:
- One sulfur atom is in the terminal position and typically has an oxidation state of -2.
- The other sulfur atom, which is bonded to the first sulfur, has an oxidation state of +6.

Thus, the oxidation states of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) are +6 and -2.

### Conclusion
The oxidation number of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) can be +6 for the central sulfur and -2 for the terminal sulfur.